From Y to Y

对于给定的一个长度为 $n$ 的字符串，初始时，我们将它视作 $n$ 个长度为 $1$ 的字符串的可重集，然后重复下列操作 $n-1$ 次： - 从这些字符串中任取两个字符串 $s,t$，将它们删除，将 $s+t$ 加入集合。注意这里的 $+$ 是指字符串拼接。 操作的成本被定义为 $\sum_{c\in\left\{\texttt{a},\texttt{b},\cdots,\texttt{z}\right\}}f\left(s,c\right)\times f\left(t,c\right)$，其中 $f\left(s,c\right)$ 是字符 $c$ 在 $s$ 中出现的次数。更通俗地说，每次操作的成本为**每个小写字母在两个字符串中出现次数的积**。 现在给定一个非负整数 $k\left(0\leqslant k\leqslant 10^5\right)$，请你构造一个不超过 $10^5$ 个字符的可重集，使得操作成本的最小值为 $k$。可以证明，这样的解一定是存在的。

From beginning till end, this message has been waiting to be conveyed. For a given unordered multiset of $ n $ lowercase English letters ("multi" means that a letter may appear more than once), we treat all letters as strings of length $ 1 $ , and repeat the following operation $ n-1 $ times: - Remove any two elements $ s $ and $ t $ from the set, and add their concatenation $ s+t $ to the set. The cost of such operation is defined to be , where $ f(s,c) $ denotes the number of times character $ c $ appears in string $ s $ . Given a non-negative integer $ k $ , construct any valid non-empty set of no more than $ 100000 $ letters, such that the minimum accumulative cost of the whole process is exactly $ k $ . It can be shown that a solution always exists.

The first and only line of input contains a non-negative integer $ k $ ( $ 0<=k<=100000 $ ) — the required minimum cost.

Output a non-empty string of no more than $ 100000 $ lowercase English letters — any multiset satisfying the requirements, concatenated to be a string. Note that the printed string doesn't need to be the final concatenated string. It only needs to represent an unordered multiset of letters.

12

abababab

3

codeforces

For the multiset {'a', 'b', 'a', 'b', 'a', 'b', 'a', 'b'}, one of the ways to complete the process is as follows: - {"ab", "a", "b", "a", "b", "a", "b"}, with a cost of $ 0 $ ; - {"aba", "b", "a", "b", "a", "b"}, with a cost of $ 1 $ ; - {"abab", "a", "b", "a", "b"}, with a cost of $ 1 $ ; - {"abab", "ab", "a", "b"}, with a cost of $ 0 $ ; - {"abab", "aba", "b"}, with a cost of $ 1 $ ; - {"abab", "abab"}, with a cost of $ 1 $ ; - {"abababab"}, with a cost of $ 8 $ . The total cost is $ 12 $ , and it can be proved to be the minimum cost of the process.